Question

"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. Unfortunately, the treatment also reduces the strength of the fabric. The breaking strength of untreated fabric is normally distributed with mean 51.8 pounds and standard deviation 2.6 pounds. The same type of fabric after treatment has normally distributed breaking strength with mean 19.1 pounds and standard deviation 2 pounds. A clothing manufacturer tests 3 specimens of each fabric. All 6 strength measurements are independent. (Round your answers to four decimal places.)

Answer 1

Since the specimens are independent, the sample mean of the breaking strengths of the 5 untreated specimens is normally distributed with mean 51.9 pounds and standard deviation 2.9/sqrt(5) pounds.

So the probability that the sample mean of the breaking strengths of the 5 untreated specimens exceeds 50 pounds is

P(Z > (50 - 51.9)/(2.9/sqrt(5))) = P(Z > -1.47) = P(Z < 1.47) = 0.9292.

b) Since the specimens are independent, the sample mean of the breaking strengths of the 5 untreated specimens is normally distributed with mean 51.9 pounds and standard deviation 2.9/sqrt(5); the sample mean of the breaking strengths of the 5 treated specimens is normally distributed with mean 20.5 pounds and standard deviation 1.9/sqrt(5) pounds.

Because the untreated and treated sample means are independent as well, the difference of the untreated minus treated sample means is normally distributed with mean 51.9 - 20.5 = 31.4 pounds and standard deviation sqrt[(2.9/sqrt(5))^2 + (1.9/sqrt(5))^2] = sqrt(2.404) pounds (recall that the variance of a sum or difference of *independent* random variables is the *sum* of their variances, so it follows that the standard deviation is the square root of the sum of the squares of their standard deviations).

So the probability that the sample mean of the breaking strengths of the 5 untreated specimens is at least 25 pounds greater than the sample mean of the breaking strengths of the 5 treated specimens is

P(difference of the untreated minus treated sample means >= 25)
= P(Z >= (25 - 31.4)/sqrt(2.404))
= P(Z >= -4.13)
= P(Z <= 4.13), which is very nearly 1.